8.6: Bad Presentations
- Discuss common mistakes in presentations
For many, the prospect of developing and delivering a business presentation rates right up there with death and taxes. Interestingly, that same mixture of fear and loathing is often felt by audience members as well. But it doesn’t have to be that way. The ability to craft a compelling story is a skill as old as the human race, and the need to communicate is as primal and potentially powerful.
For millions of years before the invention of modern technology, humans used the tools available to perpetuate traditions and culture and to document—and often rewrite—history. Do a few internet searches and immerse yourself in the Egyptian tombs; the caves of Chauvet; or El Castillo, the Temple of Kukulcan. What you’re experiencing is a feat of both artistry and communication. Although we don’t know the full significance of these early carvings and structures, there’s no doubt that these early humans captured their world view in a way that is still deeply resonant. While the tools have changed, the communication challenges—and opportunity—remain the same: to communicate an engaging and inspiring point of view.
Regardless of whether you want to change the world, build your brand, or build a billion-dollar business, effective presentation skills are essential. To quote legendary investor, philanthropist and Berkshire Hathaway chairman and CEO Warren Buffet, “If you can’t communicate and talk to other people and get across your ideas, you’re giving up your potential.” [1] As would be expected of a numbers person, Buffet has quantified his point in talks on student campuses and professional organizations. Speaking at his alma mater in 2009, Warren Buffett told Columbia Business School students that he believed learning effective communication skills could translate into 50 percent higher lifetime earnings.
Given our vibrant storytelling tradition and with so much at stake, why are there still so many bad presentations? Wouldn’t you think that modern communication technology—considering the advances in graphics and communications software alone!—would lead to more compelling presentations? Interestingly, the problem is, to some extent, the technology. It’s estimated that 30 million PowerPoint presentations are created every day, with (seemingly) a majority of presenters opting for default layouts and templates. The problem is, we’re wired for story, not bullet points. A related failure is our use of available technology.
Seth Godin has a wonderful—and instructive—rant on these points: Really Bad PowerPoint (and how to avoid it) , blaming Microsoft wizards, templates, bui -2" class="footnote"> [2]
So how do we avoid causing “death by PowerPoint”—or by whatever presentation software we use? The common denominator of presentation mistakes is that they represent a failure of communication. This failure can be attributed to two errors: too much or too little. The error of too much is generally the result of trying to use slides as a teleprompter or a substitute to a report, or, it would seem, to bludgeon the audience into submission. Of course, this tends to have an alternate effect, namely, prompting audience members to walk out or tune out, turning their attention instead to doodling or their device of choice.
What bad presentations have too little of is emotion. Presentation expert and author of the classic Presentation Zen (and 4 related books) Gar Reynolds captures the crux of the problem: “a good presentation is a mix of logic, data, emotion, and inspiration. We are usually OK with the logic and data part, but fail on the emotional and inspirational end.” [3] There’s also a hybrid too little-too much mistake, where too little substance and/or no design sensibility is — in the mind of the presenter — offset by transitions and special effects. Heed Seth Godin’s advice: “No dissolves, spins or other transitions. None.” [4]
The 10/20/30 rule, generally attributed to venture capitalist Guy Kawasaki, is a good guideline to help you achieve a “just right” balance in your presentations. Geared for entrepreneurs pitching their business, his advice is a discipline that would improve the quality—and, effectiveness—of most presentations. In brief, 10/20/30 translates to a maximum of 10 slides, a maximum of 20 minutes and a minimum of 30 point font. [5]
While this rule is a good starting point, it doesn’t overrule your audience analysis or understanding of your purpose. Sometimes, you may need more slides or have a more involved purpose—like training people in new software or presenting the results of a research study—that takes more than 30 minutes to address. In that case, go with what your audience needs and what will make your presentation most effective. The concept behind the 10/20/30 rule—to make new learning easy for your audience to take in, process and remember—should still be your guide even if you don’t follow the rule exactly.
For more on how to avoid causing death by PowerPoint, watch Swedish presentation expert and How to Avoid Death By PowerPoint author David Phillips TED Talk on the topic:
- Gallo, Carmine. " How Warren Buffet and Joel Osteen Conquered Their Terrifying Fear of Public Speaking ," Forbes . May 16, 2013.
- Rouse, Margaret. " What is death by PowerPoint? " TechTarget Network .
- Reynolds, Garr. “10 tips for Improving Your Presentations Today,” Presentation Zen. Nov 2014.
- Godin, Seth. Fix Your Really Bad PowerPoint . Ebook, sethgodin.com, 2001.
- Kawasaki, Guy. The 10/20/30 Rule of PowerPoint . December 2005.
Contributors and Attributions
- Bad Presentations. Authored by : Nina Burokas. Provided by : Lumen Learning. License : CC BY: Attribution
- Echnaton (Akhenaten) as Sphinx. Authored by : Hans Ollermann. Located at : https://flic.kr/p/4nvAVm . License : CC BY: Attribution
- How to avoid death By PowerPoint. Authored by : David JP Phillips. Provided by : TED. Located at : https://youtu.be/Iwpi1Lm6dFo . License : All Rights Reserved


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TEMPERATURE
Oral Temperature Technique
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Remove the probe from the device and place a probe cover (from the box) on the oral thermometer without touching the cover with your hands
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Place the thermometer in the mouth under the tongue in the posterior sublingual pocket (slightly off-centre) and instruct the client to keep mouth closed and not to bite on the thermometer
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Remove the thermometer when the device beeps
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Note the temperature on the digital display of the device
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Discard the probe cover in the garbage (without touching the cover)
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Place the probe back into the device
Tympanic Temperature Technique
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Remove the tympanic thermometer from the casing and place a probe cover (from the box) on the thermometer tip without touching the cover with your hands
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Turn the device on
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For an adult or older child, gently pull the helix up and back to visualize the ear canal. For an infant or younger child (under 3), gently pull the lobe down
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Gently insert the probe into the opening of the ear
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Activate the device
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Note the temperature on the digital display of the device
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Discard the probe cover in the garbage (without touching the cover) and place the device back into the holder
Axillary Temperature Technique
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Remove the probe from the device and place a probe cover (from the box) on the thermometer without touching the cover with your hands
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Ask the client to raise the arm away from his/her body
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Place the thermometer in the client’s armpit as high up as possible into the axillae on bare skin, with the point facing behind the client, and ask the client to lower arm
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Note the temperature on the digital display of the device
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Discard the probe cover in the garbage (without touching the cover) and place the probe back into the device
Rectal Temperature Technique
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Ensure the client’s privacy and wash your hands and put on gloves
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Position the client appropriately
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Remove the probe from the device and place a probe cover on it
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Lubricate the cover with a water-based lubricant
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Gently insert the probe 2–3 cm inside the rectal opening of an adult, or less depending on the size of the client
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Note the temperature on the digital display of the device when it beeps
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Discard the probe cover in the garbage (without touching the cover) and place the probe back into the device
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Remove your gloves and wash your hands
PULSE AND RESPIRATION
Radial Pulse Technique
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Use the pads of your first three fingers to gently palpate the radial pulse along the radius bone close to the flexor aspect of the wrist
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Press down with your fingers until you can best feel the pulsation
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Note the rate, rhythm, force, and equality when measuring the radial pulse
Carotid Pulse Technique
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Ask the client to sit upright
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Locate the carotid artery medial to the sternomastoid muscle in the middle third of the neck
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Gently palpate the carotid artery one at a time
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Note the rate, rhythm, force, and equality when measuring the carotid pulse
Apical Pulse Technique
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Ask the client to lay flat in a supine position
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Physically palpate the intercostal spaces to locate the landmark of the apical pulse
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Auscultate the apical pulse
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Note the rate and rhythm
Brachial Pulse Technique
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Palpate the bicep tendon in the area of the antecubital fossa
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Move your fingers medial from the tendon and about one inch above the antecubital fossa to locate the pulse
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Note the rate and rhythm
Respiration Technique
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Leave your fingers in place when you are done counting the pulse, and then begin assessing respiration
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Observe the rise and fall of the chest or abdomen
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Count for 30 seconds if the rhythm is regular or for a full minute if it is irregular
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Report respiration as breaths per minute, as well as whether breathing is relaxed, silent, and has a regular rhythm
OXYGEN SATURATION
Pulse oximter
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Remove client nail polish
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Clean oximeter probe with alcohol swab
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Clip or tape probe onto a client’s finger
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Turn oximeter on
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Take radial pulse (30 seconds if regular and one minute if irregular)
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Ensure radial pulse is aligned with pulse displayed on the oximeter
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Document or report findings
BLOOD PRESSURE
Two-step blood pressure technique
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Palpate the radial or brachial artery, inflate the blood pressure cuff until the pulse is no longer felt, and then continue to inflate 20–30 mm Hg more: this is the maximum pressure inflation.
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Deflate the cuff quickly.
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Now, you can start blood pressure so place the bell of the cleansed stethoscope over the brachial artery using a light touch and complete seal.
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Inflate the cuff to the maximum pressure inflation number.
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Open the valve slightly.
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Deflate the cuff slowly and evenly at about 2 mm Hg per second.
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Note the points at which you hear the first appearance of Korotkoff sound (systolic blood pressure), and the last Korotkoff sound before it goes silent (diastolic blood pressure).
One-step blood pressure technique
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Palpate the radial or brachial artery, inflate the blood pressure cuff until the pulse is no longer felt, and then continue to inflate 20–30 mmHg more.
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Place the bell of the cleansed stethoscope over the brachial artery using a light touch, but with an airtight seal.
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Open the valve slightly.
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Deflate the cuff slowly and evenly at about 2 mm Hg per second.
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Note the points at which you hear the first appearance of Korotkoff sound (systolic blood pressure), and the last Korotkoff sound before it goes silent (diastolic blood pressure).


13.7: Gender Communication Summary
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In this chapter you have been exposed to the specialization of gender and communication. You learned that gender communication is “the social construction of masculinity or femininity as it aligns with designated sex at birth in a specific culture and time period. Gender identity claims individuality that may or may not be expressed outwardly, and may or may not correspond to one’s sexual anatomy” (Pettitt). It is important to remember as we discuss gender and communication that there is a difference between sex and gender. Sex refers to the biological distinctions that make us male or female. Gender is the socially constructed enactment of what it means to be a man or a woman. We are generally born as either male or female, but taught how to be men and women.
People of all identities are gendered and experience their genders in a variety of ways. As a result of how gender is manifested, many feminist, men, and other activist groups have formed for the purpose of banding together with others who understand gender in similar ways. We discussed 12 types of feminism and five different men’s groups that focus on various approaches for understanding and enacting gender. There are a variety of theories that seeks to explain how we form gender. Remember that theories are simply our best representations of something. Thus theories of gender development such as Psychodynamic theory, Social Interactionism, Social Learning theory, Cognitive Learning theory, and Standpoint theory are all attempts to explain the various ways we come to understand and enact our genders.
Like with many other specializations in the field of Communication, gender communication applies to a variety of other specializations. Interpersonal communication, organizational communication, and mass communication are specializations that are particularly ripe for exploring the impact of gender and communication. Gender communication research continues to explore gender in these contexts, thus helping redefine how gender is understood and behaved.
We explored differences in gender communication styles by looking at language, the purpose of communication, patterns of talk, and nonverbal communication. While impossible to come to a definitive conclusion, gender and communication studies generally promotes the idea that the differences in gender communication are socially learned and are thus fluid and dynamic. Males and females learn to communicate in both masculine and feminine styles and make strategic choices about which style is more effective for a given context.


1.5: Boundaries
Boundaries are limits we set as individuals that define our levels of comfort when interacting with others. Personal boundaries include limits in physical, sexual, intellectual, emotional, sexual, and financial areas of our lives. Boundaries promote psychological safety in relationships at work, home, and with partners by protecting one’s well-being and limiting the stress response. For example, if you come away from a meeting or conversation with someone feeling depleted, anxious, or tense, consider if your boundaries were crossed. A lack of healthy personal boundaries can lead to emotional and physical fatigue. [1]
Five major types of personal boundaries include the following [2] :
- Physical: Physical boundaries refer to one’s personal space, privacy, and body. For example, some people are comfortable with public displays of affection (hugs, kisses, and hand-holding), while others prefer not to be touched in public.
- Sexual: Sexual boundaries refer to one’s comfort level with intimacy and attention of a sexual nature. This can include sexual comments and touch, not just sexual acts.
- Intellectual: Intellectual boundaries refer to one’s thoughts and beliefs. Intellectual boundaries are not respected when someone dismisses another person’s ideas and opinions.
- Emotional: Emotional boundaries refer to a person’s feelings. For example, an individual might not feel comfortable sharing feelings with another person and prefer to share information gradually over time.
- Financial: Financial boundaries refer to how one prefers to spend or save money.
When caring for clients with mental health disorders, it is common to notice problems with setting appropriate boundaries. For example, a client experiencing bipolar disorder may exhibit a lack of financial and sexual boundaries. When they are experiencing a manic episode, they may spend thousands of dollars on a credit card over a weekend or have sexual relations with someone they just met. Another example of boundary issues is an individual with a depressive disorder who is treated poorly by their partner but does not leave or assert boundaries because they don’t feel that they deserve to be treated any better.
Nurses must establish professional boundaries with all clients while also maintaining a respectful and caring relationship. Due to their professional role, nurses have authority and access to sensitive information that can make clients feel vulnerable. A Nurses Guide to Professional Boundaries by the National Council of State Boards of Nursing (NCSBN) states that it is the nurse’s responsibility to use clinical judgment to determine and maintain professional boundaries. Nurses should limit self-disclosure of personal information and avoid situations where they have a personal or business relationship with a client. The difference between a caring nurse-client relationship and an over-involved relationship can be difficult to discern, especially in small communities or in community health nursing where roles may overlap. In these circumstances, it is important for the nurse to openly acknowledge their dual relationship and emphasize when they are performing in a professional capacity. Signs of inappropriate boundaries include the following [3] :
- Self-disclosing intimate or personal issues with a client
- Engaging in behaviors that could be interpreted as flirting
- Keeping secrets with a client
- Believing you are the only one who truly understands or can help the client
- Spending more time than is necessary with a particular client
- Speaking poorly about colleagues or your employment setting with the client and/or their family
- Showing favoritism to a particular client
- Meeting a client in settings outside of work
- Contacting a client and/or their family members using social media
Establishing professional boundaries with clients diagnosed with mental health disorders is essential due to the vulnerability of the client population, as well as the behavioral manifestations of some disorders. For safety purposes, nurses and nursing students should keep their last name, home address, personal telephone number, and social media handles private.
View the NCSBN video: “ Professional Boundaries in Nursing .”
Read A Nurse’s Guide to Professional Boundaries PDF from the National Council of State Boards of Nursing (NCSBN)
- Pattemore, C. (2021, June 2). 10 ways to build and preserve better boundaries. PsychCentral. https://psychcentral.com/lib/10-way-to-build-and-preserve-better-boundaries#types ↵
- Pattemore, C. (2021, June 2). 10 ways to build and preserve better boundaries. PsychCentral. https://psychcentral.com/lib/10-way-to-build-and-preserve-better-boundaries#types ↵
- National Council of State Boards of Nursing. (2018). A nurse's guide to professional boundaries . https://www.ncsbn.org/ProfessionalBoundaries_Complete.pdf ↵


26.5: Fluorescent Auto-Antibody Techniques
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- Last updated
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- Ying Liu
- City College of San Francisco
Learning Objectives
- Describe the benefits of immunofluorescent antibody assays in comparison to nonfluorescent assays
- Compare direct and indirect fluorescent antibody assays
- Explain how a flow cytometer can be used to quantify specific subsets of cells present in a complex mixture of cell types
- Explain how a fluorescence-activated cell sorter can be used to separate unique types of cells
Rapid visualization of bacteria from a clinical sample such as a throat swab or sputum can be achieved through fluorescent antibody (FA) techniques that attach a fluorescent marker (fluorogen) to the constant region of an antibody, resulting in a reporter molecule that is quick to use, easy to see or measure, and able to bind to target markers with high specificity. We can also label cells, allowing us to precisely quantify particular subsets of cells or even purify these subsets for further research.
As with the enzyme assays, FA methods may be direct, in which a labeled mAb binds an antigen, or indirect, in which secondary polyclonal antibodies bind patient antibodies that react to a prepared antigen. Applications of these two methods were demonstrated in Figure 2.3.8 . FA methods are also used in automated cell counting and sorting systems to enumerate or segregate labeled subpopulations of cells in a sample.
Direct Fluorescent Antibody Techniques
Direct fluorescent antibody (DFA) tests use a fluorescently labeled mAb to bind and illuminate a target antigen. DFA tests are particularly useful for the rapid diagnosis of bacterial diseases. For example, fluorescence-labeled antibodies against Streptococcus pyogenes (group A strep) can be used to obtain a diagnosis of strep throat from a throat swab. The diagnosis is ready in a matter of minutes, and the patient can be started on antibiotics before even leaving the clinic. DFA techniques may also be used to diagnose pneumonia caused by Mycoplasma pneumoniae or Legionella pneumophila from sputum samples (Figure \(\PageIndex{1}\)). The fluorescent antibodies bind to the bacteria on a microscope slide, allowing ready detection of the bacteria using a fluorescence microscope. Thus, the DFA technique is valuable for visualizing certain bacteria that are difficult to isolate or culture from patient samples.
Indirect Fluorescent Antibody Techniques
Indirect fluorescent antibody (IFA) tests (Figure \(\PageIndex{2}\)) are used to look for antibodies in patient serum. For example, an IFA test for the diagnosis of syphilis uses T. pallidum cells isolated from a lab animal (the bacteria cannot be grown on lab media) and a smear prepared on a glass slide. Patient serum is spread over the smear and anti-treponemal antibodies, if present, are allowed to bind. The serum is washed off and a secondary antibody added. The secondary antibody is an antihuman immunoglobulin conjugated to a fluorogen. On examination, the T. pallidum bacteria will only be visible if they have been bound by the antibodies from the patient’s serum.
The IFA test for syphilis provides an important complement to the VDRL test discussed in Detecting Antigen-Antibody Complexes . The VDRL is more likely to generate false-positive reactions than the IFA test; however, the VDRL is a better test for determining whether an infection is currently active.
IFA tests are also useful for the diagnosis of autoimmune diseases. For example, systemic lupus erythematosus (SLE) (see Autoimmune Disorders ) is characterized by elevated expression levels of antinuclear antibodies (ANA). These autoantibodies can be expressed against a variety of DNA-binding proteins and even against DNA itself. Because autoimmunity is often difficult to diagnose, especially early in disease progression, testing for ANA can be a valuable clue in making a diagnosis and starting appropriate treatment.
The IFA for ANA begins by fixing cells grown in culture to a glass slide and making them permeable to antibody. The slides are then incubated with serial dilutions of serum from the patient. After incubation, the slide is washed to remove unbound proteins, and the fluorescent antibody (antihuman IgG conjugated to a fluorogen) added. After an incubation and wash, the cells can be examined for fluorescence evident around the nucleus (Figure \(\PageIndex{3}\)). The titer of ANA in the serum is determined by the highest dilution showing fluorescence. Because many healthy people express ANA, the American College of Rheumatology recommends that the titer must be at least 1:40 in the presence of symptoms involving two or more organ systems to be considered indicative of SLE. 1
Query \(\PageIndex{1}\)
Query \(\PageIndex{1}\)
Flow Cytometry
Fluorescently labeled antibodies can be used to quantify cells of a specific type in a complex mixture using flow cytometry (Figure \(\PageIndex{4}\)), an automated, cell-counting system that detects fluorescing cells as they pass through a narrow tube one cell at a time. For example, in HIV infections, it is important to know the level of CD4 T cells in the patient’s blood; if the numbers fall below 500 per μL of blood, the patient becomes more likely to acquire opportunistic infections; below 200 per μL, the patient can no longer mount a useful adaptive immune response at all. The analysis begins by incubating a mixed-cell population (e.g., white blood cells from a donor) with a fluorescently labeled mAb specific for a subpopulation of cells (e.g., anti-CD4). Some experiments look at two cell markers simultaneously by adding a different fluorogen to the appropriate mAb. The cells are then introduced to the flow cytometer through a narrow capillary that forces the cells to pass in single file. A laser is used to activate the fluorogen. The fluorescent light radiates out in all directions, so the fluorescence detector can be positioned at an angle from the incident laser light.
Figure \(\PageIndex{4}\) shows the obscuration bar in front of the forward-scatter detector that prevents laser light from hitting the detector. As a cell passes through the laser bar, the forward-scatter detector detects light scattered around the obscuration bar. The scattered light is transformed into a voltage pulse, and the cytometer counts a cell. The fluorescence from a labeled cell is detected by the side-scatter detectors. The light passes through various dichroic mirrors such that the light emitted from the fluorophore is received by the correct detector.
Data are collected from both the forward- and side-scatter detectors. One way these data can be presented is in the form of a histogram. The forward scatter is placed on the y -axis (to represent the number of cells), and the side scatter is placed on the x -axis (to represent the fluoresence of each cell). The scaling for the x -axis is logarithmic, so fluorescence intensity increases by a factor of 10 with each unit increase along the axis. Figure \(\PageIndex{5}\) depicts an example in which a culture of cells is combined with an antibody attached to a fluorophore to detect CD8 cells and then analyzed by flow cytometry. The histogram has two peaks. The peak on the left has lower fluorescence readings, representing the subset of the cell population (approximately 30 cells) that does not fluoresce; hence, they are not bound by antibody and therefore do not express CD8. The peak on the right has higher fluorescence readings, representing the subset of the cell population (approximately 100 cells) that show fluorescence; hence, they are bound by the antibody and therefore do express CD8.
Cell Sorting Using Immunofluorescence
The flow cytometer and immunofluorescence can also be modified to sort cells from a single sample into purified subpopulations of cells for research purposes. This modification of the flow cytometer is called a fluorescence-activated cell sorter (FACS) . In a FACS, fluorescence by a cell induces the device to put a charge on a droplet of the transporting fluid containing that cell. The charge is specific to the wavelength of the fluorescent light, which allows for differential sorting by those different charges. The sorting is accomplished by an electrostatic deflector that moves the charged droplet containing the cell into one collecting vessel or another. The process results in highly purified subpopulations of cells.
One limitation of a FACS is that it only works on isolated cells. Thus, the method would work in sorting white blood cells, since they exist as isolated cells. But for cells in a tissue, flow cytometry can only be applied if we can excise the tissue and separate it into single cells (using proteases to cleave cell-cell adhesion molecules) without disrupting cell integrity. This method may be used on tumors, but more often, immunohistochemistry and immunocytochemistry are used to study cells in tissues.
Query \(\PageIndex{1}\)
Table \(\PageIndex{1}\) compares the mechanisms of the fluorescent antibody techniques discussed in this section.
| Type of Assay | Mechanism | Examples |
|---|---|---|
| Direct fluorescent antibody (DFA) | Uses fluorogen-antibody conjugates to label bacteria from patient samples | Visualizing Legionella pneumophila from a throat swab |
| Indirect fluorescent antibody (IFA) | Detects disease-specific antibodies in patent serum | Diagnosing syphilis; detecting antinuclear antibodies (ANA) for lupus and other autoimmune diseases |
| Flow cytometry | Labels cell membranes with fluorogen-antibody conjugate markers excited by a laser; machine counts the cell and records the relative fluorescence | Counting the number of fluorescently labeled CD4 or CD8 cells in a sample |
| Fluorescence activated cell sorter (FACS) | Form of flow cytometry that both counts cells and physically separates them into pools of high and low fluorescence cells | Sorting cancer cells |
Key Concepts and Summary
- Immunofluorescence assays use antibody-fluorogen conjugates to illuminate antigens for easy, rapid detection.
- Direct immunofluorescence can be used to detect the presence of bacteria in clinical samples such as sputum.
- Indirect immunofluorescence detects the presence of antigen-specific antibodies in patient sera. The fluorescent antibody binds to the antigen-specific antibody rather than the antigen.
- The use of indirect immunofluorescence assays to detect antinuclear antibodies is an important tool in the diagnosis of several autoimmune diseases.
- Flow cytometry uses fluorescent mAbs against cell-membrane proteins to quantify specific subsets of cells in complex mixtures.
- Fluorescence-activated cell sorters are an extension of flow cytometry in which fluorescence intensity is used to physically separate cells into high and low fluorescence populations.
Footnotes
- 1 Gill, James M., ANNA M. Quisel, PETER V. Rocca, and DENE T. Walters. “Diagnosis of systemic lupus erythematosus.” American family physician 68, no. 11 (2003): 2179-2186.


2.10: Instantaneous Velocity
What can we say about velocity if the position graph is a curve?
What is the velocity at \(t = 5\)? Seriously, what is it? I want to know.
Well, a first stab might be that it is the slope, since that is what we said before. But generally slope is only applied to lines, not curves like this. How could we find slope for a curve?
Well, recall that slope is just how steeply something is increasing. It turns out, we can take a line that is increasing with the same steepness as the curve at a given point, and measure the slope of the line.
Here, the red line is the same steepness as the curve at \(t = 5\). This is called a tangent line . Note that often the tangent line only touches the curve once. Since it is a line, we can measure the slope, and this should represent the velocity at \(t = 5\). But since it touches one time, we don’t have two points to compute the slope. This may seem like a minor problem, but to find the exact slope takes one of the major insights in calculus: we need to develop a process to get closer and closer, and then use a limit to find the exact value . This might seem like a lot of work but it is worth it, as it demonstrates the power of calculus.
Let’s zoom in a bit on the graph:
Okay, now we can start to see some things. First, notice we know the \(y\)-value or function value at the point \(t = 5\). Why? Since the function \(p(t) = 15t - t^2\) is giving the position, we can plug in \(t = 5\). We see \(p(5) = 15(5) - (5)^2 = 50\).
But this doesn’t give us the slope. To find a slope, you need two points — then you can use the rise over run formula. But we only have one point. So instead let’s look at a line that does have two intersection points, but is not quite the line we want.
Let’s say this new green line hits the curve at \(t = 6\). A line like this green one that hits in two locations is called a secant line . What is the \(y\)-value at \(t = 6\)? Well, it’s \(p(6) = 15(6) - (6)^2 = 54\).
Now we can find the slope of the green line. It’s \({\color{green} \frac{54 - 50}{6-5} = \frac{4}{1} = 4}\), since that is what rise over run tells us. But again, it isn’t quite the line we want. Instead, we could choose a blue line that’s even closer.
Now the slope of the blue line is \({\color{blue} \frac{52.25 - 50}{5.5-5} = \frac{2.25}{0.5} = 4.5}\). Still not there. But we can get closer and closer and closer. Instead of repeating this calculation every time, let us use variables. So instead of the blue line crossing at \(t = 5.5\), let’s just say it crosses at \(t = 5 + h\), or \(h\) to the right of where the red line crosses.
What is the blue question mark? Well, we just plug \(t = 5 + h\) in to \(p(t) = 15t - t^2\). We see
\[\begin{align*} p(5 + h) & = 15(5 + h) - (5 + h)^2 \\ & = 75 + 15h - (25 + 10h + h^2) \\ & =75 + 15h - 25 - 10h - h^2 \\ & = 50 + 5h - h^2 \end{align*}\]
There you go.
Now the slope is rise over run, and we have
\[\begin{align*} {\color{blue} \text{slope} } & = {\color{blue} \frac{50 + 5h - h^2 - 50}{5 + h - 5} }\\ & = {\color{blue}\frac{5h - h^2}{h}} \end{align*}\]
Okay, what we are really interested in is when \(h\) gets really, really small. That’s when the blue line slope will equal the red line slope, and the red line slope is what we want. But we can’t plug in \(h = 0\), since that would give a division by zero explosion. So instead, we can use a limit!
\[\begin{align*} {\color{red}\text{slope}} & = \lim_{h \to 0} {\color{blue} \frac{5h - h^2}{h}} \\ & = \lim_{h \to 0} {\color{blue} \frac{h(5 - h)}{h}} \\ & = \lim_{h \to 0} {\color{blue} (5 - h)} \\ & = 5 - (0) \\ & = 5 \end{align*}\]
And there you have it. The slope of the red line is \({\color{red} 5}\)!
Now, suppose instead of \(t = 5\), we were interested in the instantaneous velocity at \(t = x\). And suppose instead of \(15t^2 - t^2\), we had some other function describing position, which we call \(f(t)\). What would the picture look like then? Well, it would look very similar to the last picture.
And this is the picture that gives us the definition of the derivative. What is the blue slope? It’s
\({\color{blue} \text{Blue Slope} = \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h}} \)
So how do we find the red slope? We just take the limit as the blue line approaches the red — that is, see what happens as \(h\) goes to zero.
\({\color{red} \text{Definition of Derivative:}} \)
\({\color{red} f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}}\)
Here is the notation for the derivative: \(f'(x)\), \(\frac{df}{dx}\), or \(\frac{d}{dx} f(x)\).


12.8.11: Procedure- Positioning Patient in Chair or Wheelchair
Positioning a patient comfortably and safely in a chair or wheelchair is an important task of the Home Health Aide/Personal Care Aide. Just as in bed, patients need to be re-positioned every two hours when they are in a chair or wheelchair. Patients can also slide down while in a wheelchair or chair, and may require assistance to maintain a proper sitting position. Patients should sit on pressure distribution devices, such as wheelchair cushions. This helps to prevent pressure ulcer formation. If the patient is able, Home Health Aides/Personal Care Aides should instruct them to shift their weight from side to side every two hours to relieve pressure on their buttocks. Use a draw sheet/bed pad under a patient sitting in a wheelchair or chair. This will help a HHA/PCA with safely repositioning and lifting the patient.
- Explain the procedure to the patient.
- Wash your hands and don gloves as needed.
- If the chair reclines : Obtain assistance from another person. Lock the wheels. Recline the chair. This allows gravity to help you move the patient up in the chair. Using a draw sheet, fold and grasp one end. Ensure you use proper body mechanics. Just as in the procedure , “ Turning the Patient in Bed ”, on the count of three, each person will lift the person up toward the head of the chair.
- If the patient is able to assist you with standing : Lock the wheelchair or chair locks. Move footrests or stools out of the way. Apply a gait belt, if using. Instruct the patient to hold onto the armrests and to push down on your instruction.  Ensure you use proper body mechanics. On the count of three, ask the person to push down against the armrests as you assist them to a stand.
- Once the patient is standing : Straighten wrinkles in chair pads or draw sheets, fluff pillows, and check their skin for any reddened areas.
- When positioning the patient back into the wheelchair or chair : Reverse the procedure used to assist them to a stand. Ensure patient’s hips are as far back in the chair as possible. Position feet so they are in proper alignment with their body. Make sure feet are on footrests if using a wheelchair and that they are not caught behind the footrests or chair legs.
- Check for positioning of male genitals. The patient should not be sitting on their scrotum.
- Place a pillow behind the patient’s back if they wish.
- Position arms on arm rests or comfortably on the patient’s lap. You may also use pillows for positioning the arms.
- If the patient requests, recline their legs and feet by reclining the chair, if it reclines, or by positioning a stool or foot rest in front of them. Elevate legs and feet with pillows, ensuring the heels are floating off the pillow. This prevents pressure ulcer formation.
- Remove gloves if wearing and wash your hands.
- Document the procedure and any observations or changes in condition.
- Just as in bed, reposition the patient every two hours when sitting in a wheelchair or chair.
- Wheelchair pillows and pressure redistribution devices should be used with patients who will spend long periods of time in wheelchairs or chairs. These will help prevent pressure ulcers.


2.8: The Simple Magnifier
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Learning Objectives
By the end of this section, you will be able to:
- Understand the optics of a simple magnifier
- Characterize the image created by a simple magnifier
The apparent size of an object perceived by the eye depends on the angle the object subtends from the eye. As shown in Figure \(\PageIndex{1}\), the object at \(A\) subtends a larger angle from the eye than when it is position at point \(B\). Thus, the object at \(A\) forms a larger image on the retina (see \(OA′\)) than when it is positioned at \(B\) (see \(OB′\)). Thus, objects that subtend large angles from the eye appear larger because they form larger images on the retina.
We have seen that, when an object is placed within a focal length of a convex lens, its image is virtual, upright, and larger than the object (see part (b) of this Figure ). Thus, when such an image produced by a convex lens serves as the object for the eye, as shown in Figure \(\PageIndex{2}\), the image on the retina is enlarged, because the image produced by the lens subtends a larger angle in the eye than does the object. A convex lens used for this purpose is called a magnifying glass or a simple magnifier .
To account for the magnification of a magnifying lens, we compare the angle subtended by the image (created by the lens) with the angle subtended by the object (viewed with no lens), as shown in Figure \(\PageIndex{1a}\). We assume that the object is situated at the near point of the eye, because this is the object distance at which the unaided eye can form the largest image on the retina. We will compare the magnified images created by a lens with this maximum image size for the unaided eye. The magnification of an image when observed by the eye is the angular magnification \(M\), which is defined by the ratio of the angle \(θ_{image}\) subtended by the image to the angle \(θ_{object}\) subtended by the object:
\[M=\dfrac{θ_{image}}{θ_{object}}. \nonumber \]
Consider the situation shown in Figure \(\PageIndex{1b}\). The magnifying lens is held a distance \(ℓ\) from the eye, and the image produced by the magnifier forms a distance \(L\) from the eye. We want to calculate the angular magnification for any arbitrary \(L\) and \(ℓ\). In the small-angle approximation, the angular size \(θ_{image}\) of the image is \(h_i/L\). The angular size \(θ_{object}\) of the object at the near point is \(θ_{object}=h_o/25\,cm\). The angular magnification is then
\[\underbrace{ M=\dfrac{θ_{image}}{θ_{object}}=\dfrac{h_i(25cm)}{Lh_o}}_{\text{angular magnification}} . \label{angular magnification} \]
Using the definition of linear magnification
\[m=−\dfrac{d_i}{d_o}=\dfrac{h_i}{h_o} \label{mag} \]
and the thin-lens equation
\[\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f} \nonumber \]
we arrive at the following expression for the angular magnification of a magnifying lens:
\[\begin{align} M&= \left(−\dfrac{d_i}{d_o}\right)\left(\dfrac{25\,cm}{L}\right) \\[4pt] &=−d_i\left(\dfrac{1}{f}−\dfrac{1}{d_i}\right)\left(\dfrac{25\,cm}{L}\right) \\[4pt] &= \left(1−\dfrac{d_i}{f}\right)\left(\dfrac{25\,cm}{L}\right) \label{eq10} \end{align} \]
From Figure \(\PageIndex{1b}\), we see that the absolute value of the image distance is \(|d_i|=L−ℓ\). Note that \(d_i<0\) because the image is virtual, so we can dispense with the absolute value by explicitly inserting the minus sign:
\[−d_i=L−ℓ. \label{eq34} \]
Inserting Equation \ref{eq34} into Equation \ref{eq10} gives us the final equation for the angular magnification of a magnifying lens:
\[M=\left(\dfrac{25\,cm}{L}\right) \left(1+\dfrac{L−ℓ}{f} \right). \label{eq12} \]
Note that all the quantities in this equation have to be expressed in centimeters. Often, we want the image to be at the near-point distance (e.g., \(L=25\,cm\)) to get maximum magnification, and we hold the magnifying lens close to the eye (\(ℓ=0\)). In this case, Equation \ref{eq12} gives
\[M=1+\dfrac{25\,cm}{f} \label{eq13} \]
which shows that the greatest magnification occurs for the lens with the shortest focal length. In addition, when the image is at the near-point distance and the lens is held close to the eye (\(ℓ=0\)), then \(L=d_i=25\,cm\) and Equation \ref{eq12} becomes
\[M=\dfrac{h_i}{h_o}=m \label{eq14} \]
where \(m\) is the linear magnification (Equation \ref{mag}) previously derived for spherical mirrors and thin lenses. Another useful situation is when the image is at infinity (\(L=\infty\)). Equation \ref{eq12} then takes the form
\[M(L=\infty)=\dfrac{25\,cm}{f}. \label{eq15} \]
The resulting magnification is simply the ratio of the near-point distance to the focal length of the magnifying lens, so a lens with a shorter focal length gives a stronger magnification. Although this magnification is smaller by 1 than the magnification obtained with the image at the near point, it provides for the most comfortable viewing conditions, because the eye is relaxed when viewing a distant object.
By comparing Equations \ref{eq13} and \ref{eq15}, we see that the range of angular magnification of a given converging lens is
\[\dfrac{25cm}{f} ≤ M ≤1+\dfrac{25cm}{f}. \nonumber \]
Example \(\PageIndex{1}\): Magnifying a Diamond
A jeweler wishes to inspect a 3.0-mm-diameter diamond with a magnifier. The diamond is held at the jeweler’s near point (25 cm), and the jeweler holds the magnifying lens close to his eye.
- What should the focal length of the magnifying lens be to see a 15-mm-diameter image of the diamond?
- What should the focal length of the magnifying lens be to obtain 10× magnification?
Strategy
We need to determine the requisite magnification of the magnifier. Because the jeweler holds the magnifying lens close to his eye, we can use Equation \ref{eq13} to find the focal length of the magnifying lens.
Solution
a. The required linear magnification is the ratio of the desired image diameter to the diamond’s actual diameter (Equation \ref{eq15}). Because the jeweler holds the magnifying lens close to his eye and the image forms at his near point, the linear magnification is the same as the angular magnification, so
\[\begin{align*} M &=m=\dfrac{h_i}{h_o}\\[4pt] &=\dfrac{15\,mm}{3.0\,mm} \\[4pt] &=5.0.\end{align*} \nonumber \]
The focal length f of the magnifying lens may be calculated by solving Equation \ref{eq13} for \(f\), which gives
\[M=1+\dfrac{25\,cm}{f} \nonumber \]
\[\begin{align*} f&=\dfrac{25\,cm}{M−1} \\[4pt] &= \dfrac{25\,cm}{5.0−1} \\[4pt] &= 6.3\,cm \end{align*} \nonumber \]
b. To get an image magnified by a factor of ten, we again solve Equation \ref{eq13} for \(f\), but this time we use \(M=10\). The result is
\[\begin{align*} f &=\dfrac{25\,cm}{M−1} \\[4pt] &=\dfrac{25\,cm}{10−1} \\[4pt] &=2.8\,cm. \end{align*} \nonumber \]
Significance
Note that a greater magnification is achieved by using a lens with a smaller focal length. We thus need to use a lens with radii of curvature that are less than a few centimeters and hold it very close to our eye. This is not very convenient. A compound microscope, explored in the following section, can overcome this drawback.


3.8: Exceptions to the Octet Rule
Returning briefly to classical Lewis diagrams. Consider the diagram shown below for the molecule SF 4 . In constructing this diagram, six valence electrons are placed around the sulfur and seven valence electrons are placed around each fluorine. As we attempt to pair these to form covalent bonds, we note that there are “too many” electrons on the sulfur! We clearly cannot form a covalent bond using three electrons, so we split one pair, move the single electrons into bonding position and form bonds with the remaining two fluorines. The “extra pair” of electrons just sits there on the sulfur and does not participate directly in the bonding.
This is an example of valence expansion . In general, elements below the second period in the periodic table (S, Se, Te, etc.) will commonly have 10 – 12 electrons in their valence shells. As in SF 4 , these electrons are not directly involved in the formation of covalent bonds, but they affect the overall reactivity of the particular molecule.
In general, all molecular compounds containing elements that appear below the second row in the periodic table are capable of valence expansion and you need to be very careful when you are drawing Lewis diagrams for these compounds. As we saw in Section 3.1 for the molecule nitrogen oxide (NO), stable molecules also exist in which atoms are not surrounded by an octet of electrons. Another example of this is the molecule BF 3 , which is shown in the following example.
Construct a Lewis diagram for the molecule BF 3 .
Solution
Boron has three valence electrons and fluorine has seven. The central atom in our structure will be boron (it is to the left of fluorine in the periodic table). Next, we draw the boron (our central atom) with its’ three electrons and place the three fluorines around the boron with the electrons arranged to form the three covalent bonds. Each of the fluorines have a full octet. The boron, however, is only surrounded by six electrons. Because of this, the boron in BF 3 is a powerful electron acceptor and forms strong complexes with electrons from other compounds. In Chapter 8 we will see that this property is called Lewis acidity and BF 3 is a very powerful Lewis acid.


15.14: Putting It Together- Integration
With a few skills and tools learned from this module, you can now create and insert multiple types of charts, tables, hyperlinks, videos, pictures, and icons into whatever Microsoft computer application you use. At the beginning of this module, the video talked about the real goal of learning all of this information, working with what you and other people create to accomplish your tasks and goals.
The variety of scenarios used throughout this module illustrated how you can use these integration skills at work. You can now utilize these skills for school or personal projects as well. Communication with coworkers, management, and others, will now be more easily accomplished through application integration skills. Here are a few instances that can call on those skills.
- You want to break up the long presentation to your work team—add in a video.
- Your boss asks for the latest financial information tables in your report—integrate an Excel table into your Word document.
- You want to record a few types of resources in your Excel spreadsheet for future reference—add in hyperlinks to websites.
- You want to send letters to a large amount of people—use your Access database to do a mail merge in Word.
Contributors and Attributions
- Putting It Together: Integration. Authored by : Sherri Pendleton. Provided by : Lumen Learning. License : CC BY: Attribution


3.6: Application- Center of Mass
Recall from single-variable calculus that for a region \(R = {(x, y) : a ≤ x ≤ b,0 ≤ y ≤ f (x)}\) in \(\mathbb{R}^2\) that represents a thin, flat plate (Figure 3.6.1), where \(f (x)\) is a continuous function on \([a,b]\), the center of mass of \(R\) has coordinates \((\bar x, \bar y)\) given by
\[ \bar x =\dfrac{M_y}{M} \text{ and }\bar y = \dfrac{M_x}{M}\]
where
\[M_x = \int_a^b \dfrac{(f(x))^2}{2}dx,\quad M_y = \int_a^b x f (x)\,dx, \quad M=\int_a^b f (x)\,dx \label{Eq3.27}\]
assuming that \(R\) has uniform density , i.e the mass of \(R\) is uniformly distributed over the region. In this case the area \(M\) of the region is considered the mass of \(R\) (the density is constant, and taken as 1 for simplicity).
In the general case where the density of a region (or lamina ) \(R\) is a continuous function \(δ = δ(x, y)\) of the coordinates \((x, y)\) of points inside \(R\) (where \(R\) can be any region in \(\mathbb{R}^2\) ) the coordinates \((\bar x,\bar y)\) of the center of mass of \(R\) are given by
\[\bar x =\dfrac{M_y}{M} \text{ and }\bar y = \dfrac{M_x}{M}\label{Eq3.28}\]
where
\[M_y = \iint\limits_R xδ(x, y)\,d A ,\quad M_x = \iint\limits_R yδ(x, y)\,d A,\quad M=\iint\limits_R δ(x, y)\,d A ,\label{Eq3.29}\]
The quantities \(M_x \text{ and }M_y\) are called the moments (or first moments ) of the region \(R\) about the \(x\)-axis and \(y\)-axis, respectively. The quantity \(M\) is the mass of the region \(R\). To see this, think of taking a small rectangle inside \(R\) with dimensions \(∆x \text{ and }∆y\) close to 0. The mass of that rectangle is approximately \(δ(x_∗, y_∗)∆x∆y\), for some point \((x_∗, y_∗)\) in that rectangle. Then the mass of \(R\) is the limit of the sums of the masses of all such rectangles inside \(R\) as the diagonals of the rectangles approach 0, which is the double integral \(\iint\limits_R δ(x, y)\,d A\).
Note that the formulas in Equation \ref{Eq3.27} represent a special case when \(δ(x, y) = 1\) throughout \(R\) in the formulas in Equation \ref{Eq3.29}.
Example 3.13: Center of Mass of a 2D Region
Find the center of mass of the region \(R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x^2 }\), if the density function at \((x, y) \text{ is }δ(x, y) = x+ y\).
Solution:
The region \(R\) is shown in Figure 3.6.2. We have
\[\nonumber \begin{align} M&=\iint\limits_R δ(x, y)\,d A \\[4pt] \nonumber &= \int_0^1 \int_0^{2x^2}(x+ y)\,d y\, dx \\[4pt] \nonumber &=\int_0^1 \left ( xy + \dfrac{y^2}{2} \Big |_{y=0}^{y=2x^2} \right ) \,dx \\[4pt] \nonumber &=\int_0^1 (2x^3 +2x^4 )\,dx \\[4pt] \nonumber &= \dfrac{x^4}{2}+\dfrac{2x^5}{5} \Big |_0^1 = \dfrac{9}{10}\\[4pt] \end{align}\]
and
\[\nonumber \begin{split} M_x &= \iint\limits_R yδ(x, y)\,d A \\[4pt] \nonumber &=\int_0^1 \int_0^{2x^2} y(x+ y)\,d y\, dx \\[4pt] \nonumber &=\int_0^1 \left ( \dfrac{xy^2}{2}+\dfrac{y^3}{3} \Big |_{y=0}^{y=2x^2} \right )\,dx \\[4pt] \nonumber &=\int_0^1 (2x^5 + \dfrac{8x^6}{ 3} )\,dx \\[4pt] \nonumber &=\dfrac{x^6}{3} + \dfrac{8x^7}{21} \Big |_0^1 = \dfrac{5}{7} \\[4pt] \end{split} \qquad \nonumber \begin{split} M_y &= \iint\limits_R xδ(x, y)\,d A \\[4pt] \nonumber &=\int_0^1 \int_0^{2x^2} x(x+ y)\,d y\, dx \\[4pt] \nonumber &=\int_0^1 \left ( x^2y+\dfrac{xy^2}{2} \Big |_{y=0}^{y=2x^2} \right )\,dx \\[4pt] \nonumber &=\int_0^1 (2x^4 + 2x^5 )\,dx \\[4pt] &=\dfrac{2x^5}{5} + \dfrac{x^6}{3} \Big |_0^1 = \dfrac{11}{15} \\[4pt] \end{split}\]
so the center of mass \((\bar x,\bar y)\) is given by
\[\nonumber \bar x =\dfrac{M_y}{M} = \dfrac{11/15}{9/10} = \dfrac{22}{27},\quad \bar y = \dfrac{M_x}{M} = \dfrac{5/7}{9/10}=\dfrac{50}{63}\]
Note how this center of mass is a little further towards the upper corner of the region \(R\) than when the density is uniform (it is easy to use the formulas in Equation \ref{Eq3.27} to show that \((\bar x,\bar y) = \left ( \dfrac{3}{ 4} , \dfrac{3}{ 5} \right ) \) in that case). This makes sense since the density function \(δ(x, y) = x + y\) increases as \((x, y)\) approaches that upper corner, where there is quite a bit of area.
In the special case where the density function \(δ(x, y)\) is a constant function on the region \(R\), the center of mass \((\bar x,\bar y)\) is called the centroid of \(R\).
The formulas for the center of mass of a region in \(\mathbb{R}^2\) can be generalized to a solid \(S\) in \(\mathbb{R}^ 3\). Let \(S\) be a solid with a continuous mass density function \(δ(x, y, z)\) at any point \((x, y, z)\) in \(S\). Then the center of mass of \(S\) has coordinates \((\bar x,\bar y,\bar z)\), where
\[\bar x = \dfrac{M_{yz}}{M},\quad \bar y = \dfrac{M_{xz}}{M},\quad \bar z= \dfrac{M_{xy}}{M},\label{Eq3.30}\]
where
\[M_{yz} = \iiint\limits_S xδ(x, y, z)\,dV, \quad M_{xz} = \iiint\limits_S yδ(x, y, z)\,dV,\quad M_{xy} = \iiint\limits_S zδ(x, y, z)\,dV ,\label{Eq3.31}\]
\[M = \iiint\limits_S δ(x, y, z)\,dV .\label{Eq3.32}\]
In this case, \(M_{yz}, M_{xz}\text{ and }M_{x y}\) are called the moments (or first moments ) of \(S\) around the \(yz\)-plane, \(xz\)-plane and \(x y\)-plane, respectively. Also, \(M\) is the mass of \(S\).
Example 3.14: Center of mass of a 3D Solid
Find the center of mass of the solid \(S = {(x, y, z) : z ≥ 0, x^2 + y^2 + z^2 ≤ a^2 }\), if the density function at \((x, y, z) \text{ is }δ(x, y, z) = 1\).
Solution:
The solid \(S\) is just the upper hemisphere inside the sphere of radius \(a\) centered at the origin (see Figure 3.6.3).
So since the density function is a constant and \(S\) is symmetric about the \(z\)-axis, then it is clear that \(\bar x = 0 \text{ and }\bar y = 0\), so we need only find \(\bar z\). We have
\[\nonumber M = \iiint\limits_S δ(x, y, z)\,dV = \iiint\limits_S 1dV = \text{ Volume}(S).\]
But since the volume of \(S\) is half the volume of the sphere of radius \(a\), which we know by Example 3.12 is \(\dfrac{4\pi a}{3}\), then \(M = \dfrac{2\pi a}{3}\). And
\[\nonumber \begin{align} M_{xy} &= \iiint\limits_S zδ(x, y, z)\,dV \\[4pt] \nonumber &=\iiint\limits_S z \,dV,\text{ which in spherical coordinates is} \\[4pt] \nonumber &=\int_0^{2\pi} \int_0^{\pi /2} \int_0^a (ρ \cos{φ})ρ^2 \sin{φ}\,dρ\, dφ\,dθ \\[4pt] \nonumber &=\int_0^{2\pi} \int_0^{\pi/2} \sin{φ} \cos{φ} \left ( \int_0^a ρ^3 \,dρ \right ) \,dφ\,dθ \\[4pt] \nonumber &= \int_0^{2\pi} \int_0^{\pi/2} \dfrac{a^4}{4} \sin{φ} \cos{φ}\,dφ\,dθ \\[4pt] \nonumber M_{xy}&=\int_0^{2\pi} \int_0^{\pi/2} \dfrac{a^4}{8} \sin{2φ}\,dφ\,dθ \quad (\text{since }\sin{2φ} = 2\sin{φ} \cos{φ}) \\[4pt] \nonumber &= \int_0^{2\pi} \left ( -\dfrac{a^4}{16} \cos{2φ} \Big |_{φ=0}^{ φ=\pi/2}\right ) \,dθ \\[4pt] \nonumber &=\int_0^{2\pi} \dfrac{a^4}{8}\,dθ \\[4pt] &=\dfrac{\pi a^4}{4}, \\[4pt] \end{align}\]
so
\[\nonumber \bar z = \dfrac{M_{xy}}{M} = \dfrac{\dfrac{\pi a^4}{4}}{\dfrac{2\pi a^3}{3}}=\dfrac{3a}{8}.\]
Thus, the center of mass of \(S\) is \((\bar x,\bar y,\bar z) = \left ( 0,0, \dfrac{3a}{8} \right ).\)


1.5: Formal Charge
- Calculate the formal charge on atoms in organic molecules.
- Understand the geometry of and the location of valence electrons in the hybrid orbitals of species with formal positive charge, formal negative charge, and organic radicals.
Prelude to a formal charge
Isolated atoms are neutral because they have equal protons and electrons. When an atom loses a a valence electron it becomes +ve species or cation, e.g., \(\ce{Na}\) after loosing an electron becomes \(\ce{Na^{+}}\), and \(\ce{Ca}\) after loosing two electron becomes \(\ce{Ca^{2+}}\). When an atom gains a valence electron, it becomes -ve species or anion, e.g., \(\ce{Cl}\) after gaining an electron becomes \(\ce{Cl^{-}}\), and \(\ce{O}\) after gaining two electrons becomes \(\ce{O^{2-}}\). Covalently bonded species can also be ions, e.g., \(\ce{NH3^{+}}\), and \(\ce{H3O^{+}}\) are cations, and \(\ce{HO^{-}}\), and \(\ce{CO2^{2-}}\) are anions. Organic compounds often become ions as intermediates in reactions.
A formal charge is assigned to an atom in a molecule based on the assumption that bonding electrons are shared equally.
How to calculate the formal charge?
The formal charge is calculated by subtracting an atom's valence electrons in a molecule from the valence electron of the isolated atom. Valence electrons of an isolated atom are equal to the first digit of the group number in the periodic table. For example, \(\ce{H}\) is in group 1 and has one valence electron shown as: \(\ce{\overset{\Large{\cdot}}{H}}\) in Lewis symbol. \(\ce{C}\) is in group 14 and has four valence electrons shown as: \(\ce{.\overset{\Large{\cdot}}{\underset{\Large{\cdot}}{C}}.}\), i.e., four unpaired dots in its Lewis symbol. \(\ce{N}\) is in group 15 and has five valence electrons shown as: \(\ce{.\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot}}{N}}.}\), i.e., five dots (one paired and three unpaired) in its Lewis symbol. Halogens are in group 17 and have seven valence electrons, e.g. \(\ce{.\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{Cl}}\!:}\), i.e., shown as three paired and one unpaired dots. Valence electrons in a molecule are assigned to an atom, assuming that the bonding electrons are equally shared, i.e., one electron to the atom per one bond. All the nonbonding electrons are assigned to the atom they are on.
The formula for calculating formal charge is:
\[Fc = Ve - (B + Nb)\nonumber \]
, where \(Fc\) is the formal charge, \(Ve\) is the valence electrons in an isolated atom, \(B\) is the number of bonds attached to the atom, and \(Nb\) is nonbonding electrons on the atom in the molecule.
What is the formal charge on \(\ce{C}\) in methane (\(\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}\)).
Solution
Isolated \(\ce{C}\) has four valance electrons, there are four bonds and no nonbonding electron in the molecule, so: \(Ve\) = 4, \(B\) = 4, \(Nb\) = 0
\(Fc = Ve - (B + Nb)\) = 4 - (4 + 0) = 0
Answer: 0 formal charge, i.e., \(\ce{C}\) in methane (\(\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{C}}{^{0}}\!\!-H}\)), where superscript over the atom (\(\ce{C}\) in this case) show the formal charge
Reactive intermediates and their formal charges
If one of the \(\ce{C-H}\) bonds in an organic compound breaks, three situations may arise, e.g., for \(\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{C}}-H}\) the situations are:
- one of the bonding electrons may remain on the \(\ce{C}\), i.e., (\(\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}-H}\)),
- none of the bonding electrons may remain on the \(\ce{C}\), i.e, (\(\ce{\small{H}-{\underset{\underset{\LARGE{H}} |}{C}}-H}\)), and
- both the bonding electrons may remain on the \(\ce{C}\), i.e., (\(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}-H}\)).
Each of these situations results in a \(\ce{C}\) with three bonds which are reactive intermediates. They are short-lived and tend to react to re-establish regular four bonds around the \(\ce{C}\). Figure \(\PageIndex{1}\) shows the structure and hybridization of the reactive carbon species. A specie with a single unshared electron on an atom is a free radical that tends to make a covalent bond by sharing the electron. A \(\ce{\small{H}-\overset{ \LARGE {\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}-H}\) is an example of a free. The (\(\ce{\small{H}-\overset{ \LARGE {\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}-H}\)), is sp 2 hybridized with three bonds and an unshared electron in its p-orbital, as illustrated in Figure \(\PageIndex{1}\)a.
What is the formal charge on \(\ce{C}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}\).
Solution
Isolated \(\ce{C}\) has four valance electrons, there are three bonds and one nonbonding electron on \(\ce{C}\) in the molecule, so: \(Ve\) = 4, \(B\) = 3, \(Nb\) = 1
\(Fc = Ve - (B + Nb)\) = 4 - (3 + 1) = 0
Answer: 0 formal charge, i.e. \(\ce{C}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}^{0}-H}\).
What is the formal charge on \(\ce{C}\) in \(\ce{\small{H}-{\underset{\underset{\LARGE{H}} |}{C}}\!-H}\).
Solution
Isolated \(\ce{C}\) has four valance electrons, there are three bonds and no nonbonding electron on \(\ce{C}\) in the molecule, so: \(Ve\) = 4, \(B\) = 3, \(Nb\) = 0
\(Fc = Ve - (B + Nb)\) = 4 - (3 + 1) = +1
Answer: +1 formal charge, i.e. (\(\ce{\small{H}-\overset{+1}{\underset{\underset{\LARGE{H}} |}{C}}-H}\)).
What is the formal charge on \(\ce{C}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}\).
Solution
Isolated \(\ce{C}\) has four valance electrons, and there are three bonds and two nonbonding electrons on \(\ce{C}\) in the molecule, so: V e = 4, B = 3, N b = 2
\(Fc = Ve - (B + Nb)\) = 4 - (3 + 2) = -1
Answer: 0 formal charge, i.e. \(\ce{\small{H}-\overset{ \LARGE {\cdot\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}^{-1}\!\!\!-H}\).
- A radical is a \(\ce{C}\) with three bonds and one unshared electron. \(\ce{C}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot}}{\underset{\underset{\LARGE{H}} |}{C}}\!-H}\) is an example of a radical that is sp 2 hybridized with three bonds and a p-orbital is partially filled by a single electron, as illustrated in Figure \(\PageIndex{1}\)a. A radical tends to make a bond by accepting an electron in its partially filled p-orbital.
- A carbocation is a \(\ce{C}\) with three bonds and no unshared electrons. A \(\ce{\small{H}-\overset{+1}{\underset{\underset{\LARGE{H}} |}{C}}-H}\) is an example of carbonation that is sp 2 hybridized with three bonds and a vacant p-orbital, as illustrated in Figure \(\PageIndex{1}\)b. A carbocation tends to make a bond by accepting a lone pair in its vacant p-orbital.
- A carbanion is a \(\ce{C}\) with three bonds and two unshared electrons. \(\ce{\small{H}-\overset{ \LARGE {\cdot\cdot}}{\underset{\underset{ \LARGE {H}} |}{C}}^{-1}\!\!\!-H}\) is an example of a carbanion that is sp 3 hybridized with three bonds and the fourth sp 3 orbital occupied by a lone pair, as illustrated in Figure \(\PageIndex{1}\)c. A carbanion tends to make a bond by donating the lone pair.
Species containing \(\ce{N}\) and \(\ce{O}\) with formal charges
\(\ce{N}\) has three bonds and one lone pair, e.g., in ammonia \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}\!-H}\) and \(\ce{O}\) has two bonds and two lone pairs, e.g., water \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{ \LARGE {H}} |}{O}}\large{:}}\) molecule. One more bond results in a cation, and one fewer bond results in anion species, as shown by solving the formal charges in the following examples.
What is the formal charge on a) \(\ce{N}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}} -H }\) and b) on \(\ce{O}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}} \large{:} }\)
Solution
a) Isolated \(\ce{N}\) has five valance electrons, there are three bonds and two nonbonding electrons (a lone pair) in the molecule, so: \(Ve\) = 5, \(B\) = 3, \(Nb\) = 2
\(Fc = Ve - (B + Nb)\) = 5 - (3 + 2) = 0
Answer: 0 formal charge, i.e., \(\ce{N}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}^{0}\!\!\!-H}\)
a) Isolated \(\ce{O}\) has six valance electrons, there are two bonds and four nonbonding electrons (two lone pairs) in the molecule, so: \(Ve\) = 6, \(B\) = 2, \(Nb\) = 4
\(Fc = Ve - (B + Nb)\) = 6 - (2 + 4) = 0
Answer: 0 formal charge, i.e., \(\ce{N}\) in \(\ce{O}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}}^{0}\!\! \large{:} }\)
What is the formal charge on a) \(\ce{N}\) in (\(\ce{ \small {H}-\overset{\overset{\LARGE{H}}|}{\underset{\underset{\LARGE{H}} |}{N}}\!-H}\)) and b) on \(\ce{O}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}}-H}\)
Solution
a) Isolated \(\ce{N}\) has five valance electrons, there are four bonds and no nonbonding electrons in the molecule, so: \(Ve\) = 5, \(B\) = 4, \(Nb\) = 0
\(Fc = Ve - (B + Nb)\) = 5 - (4 + 0) = +1
Answer: +1 formal charge, i.e., \(\ce{N}\) in \(\ce{N}\) in (\(\ce{ \small {H}-\overset{\overset{ \LARGE {H}}|}{\underset{\underset{\LARGE{H}} |}{N}}{^{+1}}\!\!\!\!-H}\))
a) Isolated \(\ce{O}\) has six valance electrons, there are three bonds and two nonbonding electrons (one lone pair) in the molecule, so: \(Ve\) = 6, \(B\) = 3, \(Nb\) = 2
\(Fc = Ve - (B + Nb)\) = 6 - (3 + 2) = +1
Answer: +1 formal charge, i.e., \(\ce{O}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{O}}^{+1}\!\!\!\!-H}\)
What is the formal charge on a) \(\ce{N}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}\large{:}}\) and b) on \(\ce{O}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}} {\underset{\LARGE{\cdot\cdot}}{O}} \large{:} }\)
Solution
a) Isolated \(\ce{N}\) has five valance electrons, there are two bonds and four nonbonding electrons (two lone pairs) in the molecule, so: \(Ve\) = 5, \(B\) = 2, \(Nb\) = 4
\(Fc = Ve - (B + Nb)\) = 5 - (2 + 4) = -1
Answer: -1 formal charge, i.e. \(\ce{N}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}}{\underset{\underset{\LARGE{H}} |}{N}}^{-1} \!\!\!\!\!\large{:} }\)
a) Isolated \(\ce{O}\) has six valance electrons, there is one bond and six nonbonding electrons (three lone pair) in the molecule, so: \(Ve\) = 6, \(B\) = 1, \(Nb\) = 6
\(Fc = Ve - (B + Nb)\) = 6 - (1 + 6) = -1
Answer: +1 formal charge, i.e., \(\ce{O}\) in \(\ce{\small{H}-\overset{\LARGE{\cdot\cdot}} {\underset{\LARGE{\cdot\cdot}}{O}} \large{:} ^{-1}}\)


1.14.8: Dealing with Death
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It is a fact that we will one day die, how will you prepare for that day?
Advance Care Planning
Advance care planning involves learning about the types of decisions that might need to be made, considering those decisions ahead of time, and then letting others know—both your family and your health care providers—about your preferences. These preferences are often put into an advance directive , a legal document that goes into effect only if you are incapacitated and unable to speak for yourself. An advance directive also allows you to express your values and desires related to end-of-life care. There are two main elements in an advance directive—a living will and a durable power of attorney for health care. A living will is a written document that helps you tell doctors how you want to be treated if you are dying or permanently unconscious and cannot make your own decisions about emergency treatment. A durable power of attorney for health care is a legal document naming a health care proxy, someone to make medical decisions for you at times when you are unable to do so.
Organ and Brain Donation
Data shows that 90% of adults support organ donation but only 60% are actually signed up as donors.
Organ donation is the act of taking healthy organs and tissues from one person and giving them to someone else. Making the decision to donate your organs is one of the most generous gifts you can give. There is no age limit for donation or to sign up. In 2021, one out of every three people who donated organs was over the age of 50. You’re never too old to make a difference — as of 2021, the oldest organ donor in the United States was 92.
While many people think that signing up to be an organ donor includes donating their brain, the purpose and the process of brain donation are different. Rather than helping to keep others alive, such as with kidney donation, brain donation helps advance scientific research. One donated brain can make a huge impact, potentially providing information for hundreds of studies.
Why donate:
- 106,008 men, women, and children are on the national transplant waiting list.
- Every 9 minutes another person is added to the transplant waiting list.
- 17 people die each day waiting for an organ transplant.
- Every donor can save 8 lives and enhance over 75 more .
 You can provide lifesaving organs to as many as eight people. Every registration counts.
Retirement and Savings
Retirement requires a lot of planning and consideration. In addition to finances, you need to think about when and where you’ll retire. Experts advise that you may need as much as 80 percent of your pre-retirement income to continue your current standard of living. The exact amount will depend on your individual needs.
As you plan, consider these important questions:
- At what age do you plan to retire?
- Can you participate in an employer’s retirement savings plan? This includes 401(k) plans and traditional pension plans.
- If you have a spouse or partner, will they retire when you do?
- Where do you plan to live when you retire? Will you downsize, rent, or own your home?
- Do you expect to work part-time?
- Will you have the same medical insurance you had while working? Will your insurance coverage change?
- Do you want to travel or pursue a costly, new hobby?
- Will you qualify for social security?
Social Security provides you with a source of income when you retire or if you can’t work due to a disability. It can also support your legal dependents (spouse, children, or parents) with benefits in the event of your death.
While you work, you pay Social Security taxes. This tax money goes into a trust fund that pays benefits to:
- Those who are currently retired
- To people with disabilities
- To the surviving spouses and children of workers who have died
Each year you work, you’ll get credits to help you become eligible for benefits when it’s time for you to retire.
Saving for retirement
Get tips for building your retirement savings in the Department of Labor’s (DOL) Top 10 Ways to Prepare for Retirement
Use a retirement calculator to find out the best age to claim your Social Security benefits.
Make a plan with the Retirement Saving worksheet


11: Nuclear Chemistry
In today’s society, the term radioactivity conjures up a variety of images. Nuclear power plants producing hydrocarbon-free energy, but with potentially deadly by-products that are difficult to store safely. Bombs that use nuclear reactions to produce devastating explosions with horrible side effects on the earth as we know it and on the surviving populations that would inhabit it. Medical technology that utilizes nuclear chemistry to peer inside living things to detect disease and the power to irradiate tissues to potentially cure these diseases. Fusion reactors that hold the promise of limitless energy with few toxic side products. Radioactivity has a colorful history and clearly presents a variety of social and scientific dilemmas. In this chapter we will introduce the basic concepts of radioactivity, nuclear equations and the processes involved in nuclear fission and nuclear fusion.
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- 11.1: Radioactivity
- Certain elements spontaneously produced a variety of particles. The three basic classes of particles were identified as “alpha”, “beta”, and “gamma” particles. Alpha particles were positive, relatively massive and were showed to be identical to the nucleus of the helium atom. Beta particles had a very small mass and were of higher energy and they carried a negative charge. Gamma particles were much more energetic, appeared to be neutral and were comparable to a high-energy photon of light.
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- 11.2: The Nuclear Equation
- To show radioactive decay in a chemical equation, you need to use atomic symbols. In the atomic symbol, the atomic number (the number of protons in the nucleus) appears as a subscript preceding the symbol for the element. The mass number appears as a superscript, also preceding the symbol.
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- 11.4: Positron Emission
- A positron, also called an antielectron, is an exotic bit of matter, or more correctly, an example of antimatter. A positron is the antimatter equivalent of an electron. It has the mass of an electron, but it has a charge of +1. Positrons are formed when a proton sheds its positive charge and becomes a neutron.


12.2: Fixed-Axis Rotation
With rigid bodies, we have to examine moments and at least the possibility of rotation along with the forces and accelerations we examined with particles. Some rigid bodies will translate but not rotate (translational systems), some will rotate but not translate (fixed axis rotation), and some will rotate and translate (general planar motion). Here we will examine rigid body rotation about a fixed axis. As the name would suggest, fixed axis rotation is the analysis of any rigid body that rotates about some axis that does not move. Many devices rotate about their center, though the objects do not need to rotate about their center point for this analysis to work.
We will again start with Newton's Second Law. Since this is a rigid body system, we include both the translational and rotational versions.
\[ \sum \vec{F} = m * \vec{a} \]
\[ \sum \vec{M} = I * \vec{\alpha} \]
By setting up free body diagrams, determining the equations of motion using Newton's Second Law, and solving for the unknowns, we can find forces based on the accelerations or vice versa.
Balanced Rotation
If the center of mass of the body is at the axis of rotation, which is known as balanced rotation, then acceleration at that point will be equal to zero. The pitching machine above is an example of a balanced rotation, and most fixed axis systems will be intentionally built to be balanced. With the acceleration of the center of mass being zero, the sum of the forces in both the \(x\) and \(y\) directions must be also be equal to zero.
\[ \sum F_x = 0 \]
\[ \sum F_y = 0 \]
In addition to the force equations, we will can also use the moment equations to solve for unknowns. In simple planar motion, this will be a single moment equation which we take about the axis of rotation or center of mass (remember that they are the same point in balanced rotation).
\[ \sum M_O = I_O * \alpha \]
Unbalanced Rotation
When the center of mass is not located on the axis of rotation, the center of mass will be accelerating and therefore forces will be exerted to cause that acceleration. In perfectly anchored systems these will be forces exerted by the bearings, though these forces can often be felt as vibrations in real systems.
The kinematics equations discussed in the previous chapter can be used to determine the acceleration of a point on a rotating body, that point being the center of mass in this case. After determining those accelerations, they can be put into force equations, most likely using the \(r\) and \(\theta\) directions.
\[ \sum F_r = m a_r \]
\[ \sum F_{\theta} = m a_{\theta} \]
Note that as the body rotates, the direction of the acceleration and the direction of the forces change. Also note that the further the center of mass is from the axis of rotation, the larger the mass. The larger the angular velocity, the larger these forces will be.
To supplement the force equations, we can use a moment equation about either the axis of rotation or the center of mass, as these are no longer the same point. Whichever is chosen, just be sure to be consistent in taking the moments and the mass moment of inertia about the same point.
\[ \sum M_O = I_O * \alpha \quad\quad \text{or} \quad\quad \sum M_G = I_G * \alpha \]
More information on how to calculate the mass moment of inertia for a body can be found in Appendix 2.
Example \(\PageIndex{1}\)
A hard drive platter 8 cm in diameter accelerates at a constant rate of 150 rad/s 2 . If the hard drive weighs a uniformly distributed 0.05 kg and we approximate the hard drive as a flat circular disc, what moment does the motor need to exert to accelerate the drive at this rate?
- Solution
Example \(\PageIndex{2}\)
The drum in a washing machine can be approximated as a cylinder 0.4 meters in diameter and 0.3 meters in height with a uniformly distributed mass of 35 kilograms when full. If we wish to achieve an acceleration of 15 rad/s 2 , what torque must the motor exert at the center of the drum?
- Solution


36.1: Sensory Processes
Skills to Develop
- Identify the general and special senses in humans
- Describe three important steps in sensory perception
- Explain the concept of just-noticeable difference in sensory perception
Senses provide information about the body and its environment. Humans have five special senses: olfaction (smell), gustation (taste), equilibrium (balance and body position), vision, and hearing. Additionally, we possess general senses, also called somatosensation, which respond to stimuli like temperature, pain, pressure, and vibration. Vestibular sensation , which is an organism’s sense of spatial orientation and balance, proprioception (position of bones, joints, and muscles), and the sense of limb position that is used to track kinesthesia (limb movement) are part of somatosensation. Although the sensory systems associated with these senses are very different, all share a common function: to convert a stimulus (such as light, or sound, or the position of the body) into an electrical signal in the nervous system. This process is called sensory transduction .
There are two broad types of cellular systems that perform sensory transduction. In one, a neuron works with a sensory receptor , a cell, or cell process that is specialized to engage with and detect a specific stimulus. Stimulation of the sensory receptor activates the associated afferent neuron, which carries information about the stimulus to the central nervous system. In the second type of sensory transduction, a sensory nerve ending responds to a stimulus in the internal or external environment: this neuron constitutes the sensory receptor. Free nerve endings can be stimulated by several different stimuli, thus showing little receptor specificity. For example, pain receptors in your gums and teeth may be stimulated by temperature changes, chemical stimulation, or pressure.
Reception
The first step in sensation is reception , which is the activation of sensory receptors by stimuli such as mechanical stimuli (being bent or squished, for example), chemicals, or temperature. The receptor can then respond to the stimuli. The region in space in which a given sensory receptor can respond to a stimulus, be it far away or in contact with the body, is that receptor’s receptive field . Think for a moment about the differences in receptive fields for the different senses. For the sense of touch, a stimulus must come into contact with body. For the sense of hearing, a stimulus can be a moderate distance away (some baleen whale sounds can propagate for many kilometers). For vision, a stimulus can be very far away; for example, the visual system perceives light from stars at enormous distances.
Transduction
The most fundamental function of a sensory system is the translation of a sensory signal to an electrical signal in the nervous system. This takes place at the sensory receptor, and the change in electrical potential that is produced is called the receptor potential . How is sensory input, such as pressure on the skin, changed to a receptor potential? In this example, a type of receptor called a mechanoreceptor (as shown in Figure \(\PageIndex{1}\)) possesses specialized membranes that respond to pressure. Disturbance of these dendrites by compressing them or bending them opens gated ion channels in the plasma membrane of the sensory neuron, changing its electrical potential. Recall that in the nervous system, a positive change of a neuron’s electrical potential (also called the membrane potential), depolarizes the neuron. Receptor potentials are graded potentials: the magnitude of these graded (receptor) potentials varies with the strength of the stimulus. If the magnitude of depolarization is sufficient (that is, if membrane potential reaches a threshold), the neuron will fire an action potential. In most cases, the correct stimulus impinging on a sensory receptor will drive membrane potential in a positive direction, although for some receptors, such as those in the visual system, this is not always the case.
Sensory receptors for different senses are very different from each other, and they are specialized according to the type of stimulus they sense: they have receptor specificity. For example, touch receptors, light receptors, and sound receptors are each activated by different stimuli. Touch receptors are not sensitive to light or sound; they are sensitive only to touch or pressure. However, stimuli may be combined at higher levels in the brain, as happens with olfaction, contributing to our sense of taste.
Encoding and Transmission of Sensory Information
Four aspects of sensory information are encoded by sensory systems: the type of stimulus, the location of the stimulus in the receptive field, the duration of the stimulus, and the relative intensity of the stimulus. Thus, action potentials transmitted over a sensory receptor’s afferent axons encode one type of stimulus, and this segregation of the senses is preserved in other sensory circuits. For example, auditory receptors transmit signals over their own dedicated system, and electrical activity in the axons of the auditory receptors will be interpreted by the brain as an auditory stimulus—a sound.
The intensity of a stimulus is often encoded in the rate of action potentials produced by the sensory receptor. Thus, an intense stimulus will produce a more rapid train of action potentials, and reducing the stimulus will likewise slow the rate of production of action potentials. A second way in which intensity is encoded is by the number of receptors activated. An intense stimulus might initiate action potentials in a large number of adjacent receptors, while a less intense stimulus might stimulate fewer receptors. Integration of sensory information begins as soon as the information is received in the CNS, and the brain will further process incoming signals.
Perception
Perception is an individual’s interpretation of a sensation. Although perception relies on the activation of sensory receptors, perception happens not at the level of the sensory receptor, but at higher levels in the nervous system, in the brain. The brain distinguishes sensory stimuli through a sensory pathway: action potentials from sensory receptors travel along neurons that are dedicated to a particular stimulus. These neurons are dedicated to that particular stimulus and synapse with particular neurons in the brain or spinal cord.
All sensory signals, except those from the olfactory system, are transmitted though the central nervous system and are routed to the thalamus and to the appropriate region of the cortex. Recall that the thalamus is a structure in the forebrain that serves as a clearinghouse and relay station for sensory (as well as motor) signals. When the sensory signal exits the thalamus, it is conducted to the specific area of the cortex (Figure \(\PageIndex{2}\)) dedicated to processing that particular sense.
How are neural signals interpreted? Interpretation of sensory signals between individuals of the same species is largely similar, owing to the inherited similarity of their nervous systems; however, there are some individual differences. A good example of this is individual tolerances to a painful stimulus, such as dental pain, which certainly differ.
 There is a one-pound difference, and one bag is twice as heavy as the other. However, would it be as easy to differentiate between a 20- and a 21-pound bag?
Question : What is the smallest detectible weight difference between a one-pound bag of rice and a larger bag? What is the smallest detectible difference between a 20-pound bag and a larger bag? In both cases, at what weights are the differences detected? This smallest detectible difference in stimuli is known as the just-noticeable difference (JND).
Background : Research background literature on JND and on Weber’s Law, a description of a proposed mathematical relationship between the overall magnitude of the stimulus and the JND. You will be testing JND of different weights of rice in bags. Choose a convenient increment that is to be stepped through while testing. For example, you could choose 10 percent increments between one and two pounds (1.1, 1.2, 1.3, 1.4, and so on) or 20 percent increments (1.2, 1.4, 1.6, and 1.8).
Hypothesis : Develop a hypothesis about JND in terms of percentage of the whole weight being tested (such as “the JND between the two small bags and between the two large bags is proportionally the same,” or “. . . is not proportionally the same.”) So, for the first hypothesis, if the JND between the one-pound bag and a larger bag is 0.2 pounds (that is, 20 percent; 1.0 pound feels the same as 1.1 pounds, but 1.0 pound feels less than 1.2 pounds), then the JND between the 20-pound bag and a larger bag will also be 20 percent. (So, 20 pounds feels the same as 22 pounds or 23 pounds, but 20 pounds feels less than 24 pounds.)
Test the hypothesis : Enlist 24 participants, and split them into two groups of 12. To set up the demonstration, assuming a 10 percent increment was selected, have the first group be the one-pound group. As a counter-balancing measure against a systematic error, however, six of the first group will compare one pound to two pounds, and step down in weight (1.0 to 2.0, 1.0 to 1.9, and so on.), while the other six will step up (1.0 to 1.1, 1.0 to 1.2, and so on). Apply the same principle to the 20-pound group (20 to 40, 20 to 38, and so on, and 20 to 22, 20 to 24, and so on). Given the large difference between 20 and 40 pounds, you may wish to use 30 pounds as your larger weight. In any case, use two weights that are easily detectable as different.
Record the observations : Record the data in a table similar to the table below. For the one-pound and 20-pound groups (base weights) record a plus sign (+) for each participant that detects a difference between the base weight and the step weight. Record a minus sign (-) for each participant that finds no difference. If one-tenth steps were not used, then replace the steps in the “Step Weight” columns with the step you are using.
| Step Weight | One pound | 20 pounds | Step Weight |
|---|---|---|---|
| 1.1 | 22 | ||
| 1.2 | 24 | ||
| 1.3 | 26 | ||
| 1.4 | 28 | ||
| 1.5 | 30 | ||
| 1.6 | 32 | ||
| 1.7 | 34 | ||
| 1.8 | 36 | ||
| 1.9 | 38 | ||
| 2.0 | 40 |
Analyze the data/report the results : What step weight did all participants find to be equal with one-pound base weight? What about the 20-pound group?
Draw a conclusion : Did the data support the hypothesis? Are the final weights proportionally the same? If not, why not? Do the findings adhere to Weber’s Law? Weber’s Law states that the concept that a just-noticeable difference in a stimulus is proportional to the magnitude of the original stimulus.
Summary
A sensory activation occurs when a physical or chemical stimulus is processed into a neural signal (sensory transduction) by a sensory receptor. Perception is an individual interpretation of a sensation and is a brain function. Humans have special senses: olfaction, gustation, equilibrium, and hearing, plus the general senses of somatosensation.
Sensory receptors are either specialized cells associated with sensory neurons or the specialized ends of sensory neurons that are a part of the peripheral nervous system, and they are used to receive information about the environment (internal or external). Each sensory receptor is modified for the type of stimulus it detects. For example, neither gustatory receptors nor auditory receptors are sensitive to light. Each sensory receptor is responsive to stimuli within a specific region in space, which is known as that receptor’s receptive field. The most fundamental function of a sensory system is the translation of a sensory signal to an electrical signal in the nervous system.
All sensory signals, except those from the olfactory system, enter the central nervous system and are routed to the thalamus. When the sensory signal exits the thalamus, it is conducted to the specific area of the cortex dedicated to processing that particular sense.
Glossary
- kinesthesia
- sense of body movement
- mechanoreceptor
- sensory receptor modified to respond to mechanical disturbance such as being bent, touch, pressure, motion, and sound
- perception
- individual interpretation of a sensation; a brain function
- proprioception
- sense of limb position; used to track kinesthesia
- reception
- receipt of a signal (such as light or sound) by sensory receptors
- receptive field
- region in space in which a stimulus can activate a given sensory receptor
- receptor potential
- membrane potential in a sensory receptor in response to detection of a stimulus
- sensory receptor
- specialized neuron or other cells associated with a neuron that is modified to receive specific sensory input
- sensory transduction
- conversion of a sensory stimulus into electrical energy in the nervous system by a change in the membrane potential
- vestibular sense
- sense of spatial orientation and balance